WHERE LIKE 子句中的 SELECT

Question

I have the following SQL query:

SELECT * FROM `test2`
WHERE text LIKE (SELECT concat(startswith,"%") from test1 ORDER BY RAND() LIMIT 1)

Essentially I want to pick one row from the test1 table, and then find all the records from the test2 table that start with the characters from the startswith column in test1.

But this query doesn't do that. Where am I going wrong?

Here the tables in questing

SELECT * FROM `test1`

startswith
==========
aaa 
bbb 
ccc 

SELECT * FROM `test2`

text
====
aaa3k123k12312p03edwqeq 
aaa12313fwefrwerw   
aaafwre3we4232  
bbb123123rwqe12e1   

Answer 1

Move the subquery, and JOIN :

SELECT text
FROM `test2`
JOIN (select startswith from test1 ORDER BY RAND() LIMIT 1) t1
    ON text LIKE concat(startswith,"%")

Answer 2

像这样在连接中使用子选择

select t2.* from test2 as t2 join (SELECT RAND() as rand, startswith from test1 ORDER BY rand LIMIT 1) as t1 where t2.txt like CONCAT(t1.startswith,"%");

Answer 3

The reason your code doesn't give you the results you expect (for example, with your sample data sometimes it returns values starting with both aaa and bbb , or you may not get all the values starting with aaa ) is that the WHERE condition is evaluated for each row of test2. That means that for each row in test2, a new, random value is compared against, this can lead (potentially) to matching all lines, or none, even if all values in test1 are the start of a value in test2 .

The way around this is to create a derived table from your subquery - so that it only gets evaluated once - and then JOIN that to test2 using the LIKE comparison you are using:

SELECT t2.*
FROM (SELECT startswith FROM test1 ORDER BY RAND() LIMIT 1) t1
JOIN test2 t2 ON t2.text LIKE CONCAT(t1.startswith, '%')

Demo on dbfiddle

Note that if startswith is "ccc", you will get no results in the output for your sample data.

Answer 4

Your code seems to be like this SELECT * FROM test2WHERE text LIKE ==========% without quotations. If you put a temporary data on your query would it still get the results? Something like

SELECT * FROM `test2`
WHERE text LIKE aaa%  -->> will not run, should have quotations

and this

SELECT * FROM `test2`
WHERE text LIKE 'aaa%'

To complete your code

SELECT * FROM `test2`
WHERE text LIKE Concat((SELECT startswith from test1 ORDER BY RAND() LIMIT 1), "%")